【LeetCode】【1.two sum】（python版）

Description：
Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路：
方法一：暴力搜索，最简单也是最容易想到的方法，两层嵌套遍历数组，计算和，时间复杂度为$O(n^2)$
方法二：两次遍历，一次遍历将数组下标和对应值放入hash表中，第二次遍历计算 target - num[i] 的值，在hash表中查找是否存在该结果，时间复杂度$O(n)$，空间复杂度$O(n)$
方法三：一次遍历，一边访问数组，计算与target的差值，如果结果不存在hash表中，就把当前数组值和下标加入hash表，如果存在直接返回结果。时间复杂度$O(n)$，空间复杂度$O(n)$

后两种方法实现如下：

class Solution(object):
    def twoSum2(self, nums, target):
        value_index = dict()
        for i in range(len(nums)):
            value_index[nums[i]] = i
        for i in range(len(nums)):
            diff = target - nums[i]
            if diff in value_index:
                return [i, value_index.get(diff)]
        return []

    def twoSum3(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        value_index = dict()
        for i in range(len(nums)):
            diff = target - nums[i]
            if diff in value_index:
                return [value_index.get(diff), i]
            value_index[nums[i]] = i
        return []